Answer:
0.04455 Hz
Explanation:
Parameters given:
Wavelength, λ = 6.5km = 6500m
Distance travelled by the wave, x = 8830km = 8830000m
Time taken, t = 8.47hours = 8.47 * 3600 = 30492 secs
First, we find the speed of the wave:
Speed, v = distance/time = x/t
v = 8830000/30492 = 289.58 m/s
Frequency, f, is given as velocity divided by wavelength:
f = v/λ
f = 289.58/6500
f = 0.04455 Hz
Answer:
I think the answer is B. amount of energy present but I'm not 100% sure
Explanation:
The strength of the electric field is 5 N/C
Explanation:
The magnitude of the electric field produced by a single-point charge is given by:
![E=\frac{kQ}{r^2}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7BkQ%7D%7Br%5E2%7D)
where
is the Coulomb's constant
Q is the magnitude of the charge
r is the distance from the charge
In this problem, we have:
is the charge producing the field
r = 100 m is the distance from the charge at which we want to calculate the field
Substituting into the equation, we find the s trength of the electric field:
![E=\frac{(8.99\cdot 10^9)(6\cdot 10^{-6})}{(100)^2}=5.4 N/C \sim 5 N/C](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B%288.99%5Ccdot%2010%5E9%29%286%5Ccdot%2010%5E%7B-6%7D%29%7D%7B%28100%29%5E2%7D%3D5.4%20N%2FC%20%5Csim%205%20N%2FC)
Learn more about electric field:
brainly.com/question/8960054
brainly.com/question/4273177
#LearnwithBrainly
Answer:
Explanation:
Energy of signal being radiated per second on all sides = 71 x 10³ J .
At a distance of 220 m it is spread over an area of 4 π x (220)² because it is spreading uniformly on all sides.
So energy crossing per unit area
= ![\frac{71\times10^3}{4 \times \pi\times(220)^2}](https://tex.z-dn.net/?f=%5Cfrac%7B71%5Ctimes10%5E3%7D%7B4%20%5Ctimes%20%5Cpi%5Ctimes%28220%29%5E2%7D)
= 11.67 x 10⁻² Wm⁻²s⁻¹.
This is the intensity of the signal.
At 2200 m this intensity will further reduce by 100 times
So there it becomes equal to
11.67 x 10⁻⁴ Wm⁻² s⁻¹.